• January 19th 2008, 03:59 AM
blertta
I need your help for these 3 exercises. Ex.1 Considering that x+y=a+b and X2+y2=a2+b2 demonstrate that xn+yn=an+bn which n is a natural number. (n is the exponent.) Ex.2 Demonstrate that the quadratic sum of 2 odd numbers is not a full quadrate number. (It means (a+b)2 Ex.3 Considering that a x b> a+b a>0 and b>0 demonstrate that a+b>4. I hope you understand me and I am sorry for my english.
• January 19th 2008, 04:37 AM
mr fantastic
Quote:

Originally Posted by blertta
I need your help for these 3 exercises. Ex.1 Considering that x+y=a+b and X2+y2=a2+b2 demonstrate that xn+yn=an+bn which n is a natural number. n is the exponent. Ex.2 Demonstrate that the quadratic sum of 2 uneven numbers is not a full quadrate number. Ex.3 Considering that a x b> a+b a>0 and b>0 demonstrate that a+b>4. I hope you understand me and I am sorry for my english.

To get you started:

Ex 1: Not sure what you're asking - perhaps you should look at Fermat's Last Theorem ....?

Ex 2: Look at Euclid's formula for generating a Pythagorean triple .....
• January 19th 2008, 05:41 AM
Soroban
Hello, blertta!

I think I understand the second problem . . .

Quote:

2) Demonstrate that the sum of the squares of two odd integers is not a square.

Consider the square of an even integer: . $(2m)^2 \:=\:4m^2$
Consider the square of an odd integer: . $(2n+1)^2 \:=\:4(n^2+n) + 1$

$\text{The square of an integer is: }\begin{array}{c}\text{[1] a multiple of 4, or} \\ \text{[2] one more than a multiple of 4}\end{array}$

Now consider the sum of the squares of two odd integers:

. . $(2a+1)^2 + (2b+1)^2 \;=\;4a^2 + 4a + 1 + 4b^2 + 4b + 1 \;=\;4(a^2+b^2+a+b) + 2$

This is two more than a multiple of 4 . . . It cannot be a square.

• January 19th 2008, 09:24 AM
topsquark
Quote:

Originally Posted by blertta
Ex.3 Considering that a x b> a+b a>0 and b>0 demonstrate that a+b>4. I hope you understand me and I am sorry for my english.

There is a problem with this:

If $a = b = 0.1$ then $0.1^2 = 0.01 \not > 0.2$.

-Dan
• January 19th 2008, 01:48 PM
mr fantastic
Quote:

Originally Posted by topsquark
There is a problem with this:

If $a = b = 0.1$ then $0.1^2 = 0.01 \not > 0.2$.

-Dan

I think the intent of the question is:

IF $a b > a + b$ then a + b > 4.

In other words, show that $a b > a + b \Rightarrow a + b > 4$.

Your counter-example doesn't satisfy the necessary condition. (And yes, I too fiddled around for a bit thinking a counter-example might exist).

Perhaps the fact that $a b > a + b$ is only true when $a b - a > b \Rightarrow a(b - 1) > b$ $\Rightarrow \left\{ \begin{array}{cl} a > \frac{b}{b - 1}, & b > 1 \\ \\ a < \frac{b}{b - 1}, & 0 < b < 1 \\ \\ \text{False}, & b = 1 \end{array}\right.$

For b > 1, this gives $a + b > \frac{b}{b - 1} + b = b + 1 + \frac{1}{b - 1}$. The graph of $y = b + 1 + \frac{1}{b - 1}$ has a minimum turning point at b = 2 (which satisfies b > 1), corresponding to y = 4 ......